Download Chapter 1: Topology

c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011
Chapter 1
Topology
We start by defining a topological space.
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A topological space is a set S together with a collection O of
subsets called open sets such that the following are true:
i) the empty set ∅ and S are open, ∅, S ∈ O
ii) the intersection of a finite number of open sets is open; if
U1 , U2 ∈ O, then U1 ∩ U2 ∈ O
iii) S
the union of any number of open sets is open, if Ui ∈ O, then
Ui ∈ O irrespective of the range of i.
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It is the pair {S, O} which is, precisely speaking, a topological
space, or a space with topology. But it is common to refer to S as
a topological space which has been given a topology by specifying
O.
Example: S = R, the real line, with the open sets being open
intervals ]a, b[ , i.e. the sets {x ∈ R | a < x < b} and their unions,
plus ∅ and R itself. Then (i) above is true by definition.
For two such open sets U1 = ]a1 , b1 [ and U2 = ]a2 , b2 [ , we can
suppose a1 < a2 . Then if b1 6 a2 , the intersection U1 ∩ U2 = ∅ ∈ O .
Otherwise U1 ∩ U2 = ]a2 , b1 [ which is an open interval and thus
U1 ∩ U2 ∈ O. So (ii) is true.
And (iii) is also true by definition.
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Similarly Rn can be given a topology via open rectangles, i.e.
via the sets {(x1 , · · · , xn ) ∈ Rn | ai < xi < bi }. This is called the
standard or usual topology of Rn .
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The trivial topology on S consists of O = {∅, S}.
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c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011
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Chapter 1. Topology
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The discrete topology on a set S is defined by O = {A | A ⊂ S},
i.e., O consists of all subsets of S.
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A set A is closed if its complement in S, also written S\A or
as A{ , is open.
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Closed rectangles in Rn are closed sets as are closed balls and
single point sets.
A set can be neither open nor closed, or both open and closed.
In a discrete topology, every set A ⊂ S is both open and closed,
whereas in a trivial topology, any set A 6= ∅ or S is neither open nor
closed.
The collection C of closed sets in a topological space S satisfy
the following:
i) the empty set ∅ and S are open, ∅, S ∈ C
ii) the union of a finite number of open sets is open; if A1 , A2 ∈ C ,
then A1 ∪ A2 ∈ C
iii) the intersection
of any number of open sets is open, if Ai ∈ C ,
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then Ai ∈ C irrespective of the range of i.
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Closed sets can also be used to define a topology. Given a set S
with a collection C of subsets satisfying the above three properties of
closed sets, we can always define a topology, since the complements
of closed sets are open. (Exercise!)
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An open neighbourhood of a point P in a topological space
S is an open set containing P . A neighbourhood of P is a set
containing an open neighbourhood of P . Neighbourhoods can be
defined for sets as well in a similar fashion.
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Examples: For a point x ∈ R, and for any > 0,
]x − , x + [ is an open neighbourhood of x,
[x − , x + [ is a neighbourhood of x,
{x − 6 y < ∞} is a neighbourhood of x,
[x, x + [ is not a neighbourhood of x.
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A topological space is Hausdorff if two distinct points have
disjoint neighbourhoods.
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Topology is useful to us in defining continuity of maps.
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A map f : S1 → S2 is continuous if given any open set U ⊂ S2
its inverse image (or pre-image, what it is an image of) f −1 (U ) ⊂ S1
is open.
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When this definition is applied to functions from R to R , it is
c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011
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the same as the usual − δ definition of continuity, which says that
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f : Rm → Rn is continuous at x0 if given > 0, we can always
find a δ > 0 such that |f (x) − f (x0 )| < whenever |x − x0 | < δ. 2
For the case of functions from a topological space to Rn , this
definition says that
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f : S → Rn is continuous at s0 ∈ S if given > 0, we can
always find an open neighbourhoood U of s0 such that |f (s)−f (s0 )| <
whenever s ∈ U.
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If a map f : S1 → S2 is one-to-one and onto, i.e. a bijection,
and both f and f −1 are continuous, f is called a homeomorphism
and we say that S1 and cs2 are homeomorphic.
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Proposition: The composition of two continuous maps is a continuous map.
Proof: If f : S1 → S2 and g : S2 → S3 are continuous maps, and
U is some open set in S3 , then its pre-image g −1 (U ) is open in S2 .
So f −1 (g −1 (U )), which is the pre-image of that, is open in S1 . Thus
(g ◦ f )−1 (U ) = f −1 (g −1 (U )) is open in S1 . Thus g ◦ f is continuous.
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